传送门:A Bug's Life

Description

Background
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.         Problem Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.      

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.      

Output

The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.      

Sample Input

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output

Scenario #1:
Suspicious bugs found! Scenario #2:
No suspicious bugs found!

Hint

Huge input,scanf is recommended.
写这道题是为了说明一件重要的事:
不能认为Input总是以EOF结尾。
WA到死的写法(注意加粗的两行)
using namespace std;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<functional>
#include<vector>
#include<map>
#include<queue>
#include<climits>
#define pq priority_queue
#define X first
#define Y second
#define MP make_pair
#define pb push_back
#define Read(x) freopen(x, "r", stdin)
#define scf(x) scanf(x)
#define prf(x) printf(x)
#define set_0(x) memset(x, 0, sizeof(x))
#define set_1(x) memset(x, -1, sizeof(x));
#define rep(i, l, r) for(int i=l; i<r; i++)
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpii; const int MAX_N=2e3+;
int par[MAX_N];
bool rel[MAX_N]; int find(int x){
if(par[x]==x) return x;
int tmp=par[x];
par[x]=find(par[x]);
if(rel[x]) rel[x]=rel[tmp];
else rel[x]=!rel[tmp];
return par[x];
} void unite(int x, int y){
int yy=find(y), xx=find(x);
if(xx==yy) return;
par[xx]=yy;
if(rel[x]==rel[y]) rel[xx]=false;
else rel[xx]=true;
} int main(){
//Read("in");
scanf("%*d");
int N, M, a, b, cs=;
bool ok;
while(~scanf("%d%d", &N, &M)){
for(int i=; i<=N; i++){
par[i]=i;
rel[i]=true; //i与par[i]是否同性
}
ok=true;
while(M--){
scanf("%d%d", &a, &b);
if(!ok) continue;
if(find(a)==find(b)){
if(rel[a]==rel[b]) ok=false;
}
else unite(a, b);
}
if(cs) puts("");
printf("Scenario #%d:\n", ++cs);
puts(ok?"No suspicious bugs found!":"Suspicious bugs found!");
}
return ;
}

AC的姿势:

using namespace std;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<functional>
#include<vector>
#include<map>
#include<queue>
#include<climits>
#define pq priority_queue
#define X first
#define Y second
#define MP make_pair
#define pb push_back
#define Read(x) freopen(x, "r", stdin)
#define scf(x) scanf(x)
#define prf(x) printf(x)
#define set_0(x) memset(x, 0, sizeof(x))
#define set_1(x) memset(x, -1, sizeof(x));
#define rep(i, l, r) for(int i=l; i<r; i++)
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpii; const int MAX_N=2e3+;
int par[MAX_N];
bool rel[MAX_N]; int find(int x){
if(par[x]==x) return x;
int tmp=par[x];
par[x]=find(par[x]);
if(rel[x]) rel[x]=rel[tmp];
else rel[x]=!rel[tmp];
return par[x];
} void unite(int x, int y){
int yy=find(y), xx=find(x);
if(xx==yy) return;
par[xx]=yy;
if(rel[x]==rel[y]) rel[xx]=false;
else rel[xx]=true;
} int main(){
//Read("in");
int N, M, a, b, cs=, T;
bool ok;
scanf("%d", &T);
while(T--
){
scanf("%d%d", &N, &M);
for(int i=; i<=N; i++){
par[i]=i;
rel[i]=true; //i与par[i]是否同性
}
ok=true;
while(M--){
scanf("%d%d", &a, &b);
if(!ok) continue;
if(find(a)==find(b)){
if(rel[a]==rel[b]) ok=false;
}
else unite(a, b);
}
if(cs) puts("");
printf("Scenario #%d:\n", ++cs);
puts(ok?"No suspicious bugs found!":"Suspicious bugs found!");
}
return ;
}
 

POJ 2492 A Bug's Life的更多相关文章

  1. hdu - 1829 A Bug's Life (并查集)&&poj - 2492 A Bug's Life && poj 1703 Find them, Catch them

    http://acm.hdu.edu.cn/showproblem.php?pid=1829 http://poj.org/problem?id=2492 臭虫有两种性别,并且只有异性相吸,给定n条臭 ...

  2. POJ 2492 A Bug's Life(并查集)

    http://poj.org/problem?id=2492 题意 :就是给你n条虫子,m对关系,每一对关系的双方都是异性的,让你找出有没有是同性恋的. 思路 :这个题跟POJ1703其实差不多,也是 ...

  3. (简单) POJ 2492 A Bug's Life,二分染色。

    Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs ...

  4. POJ 2492 A Bug's Life(带权并查集)

    题目链接:http://poj.org/problem?id=2492 题目大意:有n只虫子,m对关系,m行每行有x y两个编号的虫子,告诉你每对x和y都为异性,先说的是对的,如果后面给出关系与前面的 ...

  5. POJ 2492 A Bug's Life (并查集)

    A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 30130   Accepted: 9869 De ...

  6. POJ 2492 A Bug's Life (并查集)

    Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes ...

  7. POJ 2492 A Bug's Life【并查集高级应用+类似食物链】

    Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes ...

  8. hdu 1829 &amp;poj 2492 A Bug&#39;s Life(推断二分图、带权并查集)

    A Bug's Life Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) To ...

  9. poj 2492 A Bug's Life 二分图染色 || 种类并查集

    题目链接 题意 有一种\(bug\),所有的交往只在异性间发生.现给出所有的交往列表,问是否有可疑的\(bug\)(进行同性交往). 思路 法一:种类并查集 参考:https://www.2cto.c ...

随机推荐

  1. lecture12-玻尔兹曼机和受限玻尔兹曼机

    这是Hinton的第12课,结合前一课可以知道RBM是来自BM,而BM是来自Hopfield的,因为水平有限,是直译的,虽然有时候会看不懂,但是好歹不会曲解原来的本意,看的话:1.先看ppt:2.通读 ...

  2. PHP学习资料分享

    PHP有着开源.执行效率高.优化代码运行等特性,功能强大,被广泛运用,很多大型企业网站开发的首选,百度.腾讯.新浪.金山等均采用了PHP开发.学好PHP对网络开发人员而言具有很好的发展前景,市场对PH ...

  3. JavaScript函数劫持

    一.为什么我会写这篇文章 这篇文章其实是在一个偶然的机会下发现了居然有JavaScript劫持这种东西,虽然这种东西在平时用的比较少,而且一般实用价值不高,但是在一些特殊的情况下还是要使用到的,所以在 ...

  4. Oracle中序列(SEQUENCE)的使用一例

    曾经在触发器中使用序列(SEQUENCE): create or replace trigger TRI_SUPPLIER before insert on SUPPLIER for each row ...

  5. 重叠(Overlapping) NAT

    当内部网络也使用公网注册地址(或者是外网合法地址)时,如果仍使用标准的静态或者动态NAT转换,则可能使得转换的内网地址与外网中合法地址冲突,使数据包又返回到了本地网络,这肯定是不行的.这时我们就要使用 ...

  6. java中的static详解

    如果一个类成员被声明为static,它就能够在类的任何对象创建之前被访问,而不必引用任何对象.static 成员的最常见的例子是main( ) .因为在程序开始执行时必须调用main() ,所以它被声 ...

  7. fir2(n,f,m)

    编辑 函数fir2用来设计多通带任意响应FIR滤波器,该滤波器的幅频特性由向量对f和m确定,f为归一化频率向量,m为对应频率点上的幅度.当设计的滤波器在频率为π的幅度响应不是0时,滤波器的阶数n为偶数

  8. app状态监听广播

    手机中的应用在安装.更新和卸载时都会发送广播 清单文件 <?xml version="1.0" encoding="utf-8"?> <man ...

  9. AlertDialog之常见对话框(单选对话框、多选对话框、进度条对话框)

    单选对话框,顾名思义就是只能选一项(setSingleChoiceItems(Items,)) public void click(View v){ //创建对话框类 AlertDialog.Buil ...

  10. 【JavaEE企业应用实战学习记录】servlet3.0上传文件

    <%-- Created by IntelliJ IDEA. User: Administrator Date: 2016/10/6 Time: 14:20 To change this tem ...